E2 Reactions

This page under construction.

This page covers "bimolecular elimination" reactions. The purpose of this page is to illustrate this important class of reactions, which the reader should compare with the related E1 and SN1 reactions.

The term E2 stands for "elimination bimolecular." Like any elimination reaction, the product of an E2 elimination reaction has one more degree of unsaturation than the starting materials did. For instance, the base-induced elimination of "HX" (dehydrohalogenation) of an alkyl halide gives rise to an alkene (illustrated below for the conversion of tert- butyl bromide to isobutylene).

E2 eliminations, in contrast to E1 reactions are promoted by strong base. The base vital to the reaction; it is directly involved in the rate-determining step. The reaction is bimolecular--that is, it involves "second-order kinetics--because two molecules must come together for the reaction to occur. The mechanism of an E2 elimination reaction is shown below:

Notice that the hydrogen that is removed is on the carbon atom that is adjacent to the one bearing the halogen. For some reason, beginning students are often confused on this point, although it is mysterious as to why they should be. Carbon-carbon double bonds, by definition,exist between two adjacent carbon atoms. Likewise, the "H" and the "X" atoms that are eliminated during the dehydrohalogenation of an alkyl halide must be on the carbon atoms.

Self-test question #1

There are two elimination products that could be formed by the loss of HBr from t-pentyl bromide. Can you draw them?

In the Web page on "Nomenclature," you learned about (or reviewed) the trivial names for twelve alkyl groups having five or fewer carbons. Consider the bromides derived from each of those groups. Which two of them could not possibly undergo an E2 elimination reaction, and why?

Stereochemical evidence indicates that E2 reactions always occur via "periplanar" geometry, that is, the atoms of the H-C-C-X group involved in the reaction must all lie in the same plane. This gives rise to two possible orientations:

Of these two, the "anti periplanar" geometry is obviously the most favorable.

Self-Test question #4

Consider cis- and trans-1-tert-butyl-4-chlorohexane

a)One of these molecules cannot react by an E2 pathway. Which one is it, and why can't it?

b)What is the product of the E2 elimination reaction from the isomer that can react by E2?